A 1920.-nF parallel plate capacitor is connected to a 13.4-V battery and charged

Question

A 1920.-nF parallel plate capacitor is connected to a 13.4-V battery and charged. a)What is the charge Q on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor?
The 1920.-nF capacitor is then disconnected from the 13.4-V battery and used to charge three uncharged capacitors, a 140.-nF capacitor, a 250.-nF capacitor, and a 305.-nF capacitor, connected in series.
c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the 1920.-nF capacitor was transferred to the other three capacitors?

Explanation / Answer

C0 = 3730 nF, V = 21.4 V, C1 = 150 nF, C2 = 235 nF, C3 = 345 nF
Q = C0*V
c) After charging, what is the potential difference across each of the four capacitors?

C' = (C1-1 + C2-1 + C3-1)-1 = 72.356 nF

potential difference on C0 and C' is V'

C0*V' + C'V' = Q

V' = Q/(C0 + C') = C0*V/(C0 + C') = 21.0 V

charge on C' is Q' = C'*V' = 1519 nC

potential difference across C1 is Q'/C1 = 10.1 V

potential difference across C2 is Q'/C2 = 6.5 V

potential difference across C3 is Q'/C3 = 4.4 V

d) How much of the electrical energy stored in the 3730.-nF capacitor was transferred to the other three capacitors?

the electrical energy stored in the 3730.-nF capacitor = C0*V2/2 = 854095 nJ

the electrical energy stored in the other three capacitors Q'2/(2C') = 15945 nJ

energy transferred = 854095 - 15945 = 838150 nJ = 838.15 J

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